How many grams of Al form from 9.73 g of aluminum oxide if the yield is 91\(\% ?\) $$\mathrm{Al}_{2} \mathrm{O}_{3}+3 \mathrm{C} \rightarrow 2 \mathrm{Al}+3 \mathrm{CO}$$

Short Answer

Expert verified
The actual yield of Al produced is approximately 4.67g

Step by step solution

01

Determine the Molar Masses

First, calculate the molar mass of aluminum oxide (Al2O3) by taking 2 times the atomic mass of Al plus 3 times the atomic mass of O. The atomic mass of Al is approximately 26.98 g/mol, and that of O is approximately 16 g/mol. Therefore, the molar mass of Al2O3 is \(2(26.98g/mol) + 3(16g/mol) = 101.96 g/mol\)
02

Calculate the number of moles of Al2O3

Next, calculate the number of moles of aluminum oxide in the given mass by using the molar mass. The number of mole is given by \(Mass/Molar mass = 9.73g/ 101.96 g/mol = 0.095 mol of Al2O3\)
03

Determine the Theoretical Yield of Al

From the balanced chemical equation, it can be seen that 2 moles of Al are produced from 1 mol of Al2O3. So, theoretically, the moles of Al produced will be \(=0.095 mol of Al2O3 x 2Al/1Al2O3 = 0.19mol of Al\). The mass of Al can be determined by multiplying the moles of Al by its molar mass. \(\approx 0.19 molof Al x 26.98g/mol = 5.13g\)
04

Calculate the Actual Yield

The exercise states that the yield is 91%. So, the actual amount of Al produced can be calculated by multiplying the theoretical yield by the given percentage. Thus, the actual yield is \(\approx 5.13g x 0.91 = 4.67g\)

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Describe the relationship between the limiting reactant and the theoretical yield.

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